AP EAMCET · Maths · Application of Derivatives
If \(x\) and \(y\) are two positive numbers such that \(x+y=32\), then the minimum value of \(x^2+y^2\) is,
- A 500
- B 256
- C 1024
- D 512
Answer & Solution
Correct Answer
(D) 512
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { Let, } s=x^2+y^2 \\ & =x^2+(32-x)^2 \quad[\because x+y=32] \\ & \therefore \quad \frac{d s}{d x}=2 x+2(32-x)(-1) \\ & =2 x-2(32-x) \\ & =2 x-64+2 x \\ & =4 x-64 \\ & \end{aligned} \] For maxima or minima,…
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